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Determine for what values of $x \in \Bbb R$ the series

$$\sum_{n = 1}^\infty \frac{(-1)^n}{2n+1}\left(\frac{1-x}{1+x}\right)^n$$

coverges. I have tried the alternating series test but I don't think I am doing it correctly because I keep getting infinity. Does that mean it converges for all values? Thank you.

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    $\begingroup$ would the root test apply to this situation? could I apply the root test $\endgroup$ – user226131 Mar 24 '15 at 19:39
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    $\begingroup$ That's a good idea, but I think it'll be easier for you if you apply the ratio test. $\endgroup$ – kobe Mar 24 '15 at 19:41
  • $\begingroup$ Yes you can apply the root test. $\endgroup$ – science Mar 24 '15 at 19:57
  • $\begingroup$ $\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}~=~\arctan x$. $\endgroup$ – Lucian Mar 24 '15 at 20:33
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Since $(1 - x)/(1 + x)$ is defined only for $x \neq -1$, we can rule out $x = -1$. Let $a_n(x)$ be the $n$th term of the series. Then

$$\lim_{n\to \infty} \left|\frac{a_{n+1}(x)}{a_n(x)}\right| = \lim_{n\to \infty} \frac{2n+1}{2n+3}\left|\frac{1-x}{1+x}\right| = \left|\frac{1 - x}{1 + x}\right|.$$

Now $|(1 - x)/(1 + x)| < 1 \iff|1 - x| < |1 + x| \iff|1 - x|^2 < |1 + x|^2$. By expanding both sides of the equality, we get the equivalent inequality $x > 0$. By the ratio test, the series $\sum a_n(x)$ converges when $x > 0$ and diverges when $x < 0$. When $x = 0$, the alternating series test shows $\sum a_n(x)$ converges. Thus $\sum a_n(x)$ converges if and only if $x \ge 0$.

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Applying the ratio test, one finds the limiting ratio: $$ \left|\frac{1-x}{1+x}\right|. $$ The series therefore converges if that ratio is $<1$ and diverges if it is $>1$. What happens when it is equal to $1$ must be looked at separately; the ratio test doesn't help there. So we have $$ -1<\frac{1-x}{1+x}<1. $$ We cannot multiply all three members by $1+x$ because that is sometimes positive and sometimes negative, depending on $x$. So use a common denominator: $$ \frac{-1-x}{1+x} < \frac{1-x}{1+x} < \frac{1+x}{1+x} $$ The first inequality becomes $$ 0 < \frac{2}{1+x} $$ and the second becomes $$ \frac{2x}{1+x}>0. $$ The first is satisfied if $x>-1$ and the second if either $x>0$ or $x<-1$. You need both, so the solution is $x>0$.

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